Let $M$ and $N$ be the midpoints of segments $BC$ and $AC$ of a triangle $ABC$, respectively. Let $Q$ be a point on the line through $N$ parallel to $BC$ such that $Q$ and $C$ are on opposite sides of $AB$ and $\vert QN\vert \cdot \vert BC\vert=\vert AB\vert \cdot \vert AC\vert$. Suppose that the circumcircle of triangle $AQN$ intersects the segment $MN$ a second time in a point $T \ne N$. Prove that there is a circle through points $T$ and $N$ touching both the side $BC$ and the incircle of triangle $ABC$.
Problem
Source: Germany 2022, Problem 3
Tags: geometry, circumcircle, incircle, touching circles, sidelengths, geometry proposed
alchemyst_
25.06.2022 20:06
WLOG $AB \leq AC$. Let the incircle of $\bigtriangleup{ABC}$ touch $BC$ at $D$ and let $BC = a$, $CA = b$, $AB = c$. Suppose the circle tangent to $BC$ at $D$ that passes through $N$ meets $MN$ at $T'$ other than $N$. It would suffice to show $T = T'$ as this circle will meet the conditions of the problem. By PoP, $MD^2 = MT' \times MN \Rightarrow \left(\frac{MD}{MN}\right)^2 = \frac{MT'}{MN}$. $MD = \frac{BC}{2} - BD = \frac{a}{2} - (s - b) = \frac{a - (a - b + c)}{2} = \frac{b - c}{2} \Rightarrow \frac{MT'}{MN} = \left(\frac{\frac{b - c}{2}}{\frac{c}{2}}\right)^2 = \left(\frac{b - c}{c}\right)^2 = \frac{b^2 - 2bc + c^2}{c^2} \Rightarrow \frac{MT'}{T'N} = \frac{MT'}{MN - MT'} = \frac{b^2 - 2bc + c^2}{c^2 - (b^2 - 2bc + c^2)} = \frac{b^2 - 2bc + c^2}{2bc - b^2}$. If we take a homothety centered at $C$ with scale factor $2$ then $N \rightarrow A$ and let $A \rightarrow A'$ and $T' \rightarrow L$. Now we take barycentric coordinates with reference to $\bigtriangleup{ABC}$. $A = (1, 0, 0)$ $A' = (2, 0, -1)$ $T' = \left(\frac{b^2 - 2bc + c^2}{c^2}, \frac{2bc - b^2}{c^2}, 0\right)$ Using the formula for a circle we have: $-a^2yz - b^2xz - c^2xy + (ux + vy + wz)(x + y + z) = 0$ The point $(1, 0, 0)$ gives $u = 0$. $(2, 0, -1)$ gives: $-b^2(2)(-1) + (w(-1))(1) = 0 \Rightarrow w = 2b^2$ $\left(\frac{b^2 - 2bc + c^2}{c^2}, \frac{2bc - b^2}{c^2}, 0\right)$ gives: $-c^2(b^2 - 2bc + c^2)(2bc - b^2) + (v(2bc - b^2))c^2 = 0 \Rightarrow -(b^2 - 2bc + c^2) + v = 0 \Rightarrow v = b^2 - 2bc + c^2$. If we let this circle intersect the line through $A$ parallel to $BC$ at $K \neq A$ then we will have $K = (1, y, -y)$ and: $-a^2y(-y) - b^2(-y) - c^2y + ((b^2 - 2bc + c^2)y + 2b^2(-y)) = 0, y \neq 0 \Rightarrow a^2y + b^2 - c^2 + b^2 - 2bc + c^2 - 2b^2 = 0 \Rightarrow a^2y = 2bc \Rightarrow y = \frac{2bc}{a^2} \Rightarrow K = \left(1, \frac{2bc}{a^2}, -\frac{2bc}{a^2}\right)$ We have that $\overrightarrow{AK} = \left(0, \frac{2bc}{a^2}, -\frac{2bc}{a^2}\right)$ and using distance formula we have $AK^2 = -a^2\times\frac{2bc}{a^2}\times -\frac{2bc}{a^2} = \frac{4b^2c^2}{a^2} \Rightarrow AK = \frac{2bc}{a}.$ So if we take a homothety centered at $C$ with scale factor $\frac{1}{2}$ and let $K \rightarrow Q'$ we will find that $NQ' = \frac{AK}{2} = \frac{bc}{a}$ and $Q'$ will be on the line through $N$ parallel to $BC$. Thus $Q'N \cdot BC = AB \cdot AC$ so $Q = Q'$ so $QANT'$ lie on a circle implying $T = T'$ so we are done.